Question

\[\int\frac{2 \sin x + 3 \cos x}{3 \sin x + 4 \cos x} dx\]

Answer 1

\[\text{ Let I }= \int\left( \frac{2 \sin x + 3 \cos x}{3 \sin x + 4 \cos x} \right)dx\]
\[\text{ and  let 2   sin x + 3 cos x} = A \left( 3 \sin x + 4 \cos x \right) + B \left( 3 \cos x - 4 \sin x \right) . . . (1)\]
\[ \Rightarrow 2 \sin x + 3 \cos x = \left( 3A - 4B \right) \sin x + \left( 4A + 3B \right) \cos x\]

By comparing the coefficients of like terms we get,

\[3A - 4B = 2 . . . \left( 2 \right)\]
\[4A - 3B = 3 . . . \left( 3 \right)\]

Multiplying eq (2) by 3 and eq (3) by 4 and then adding,

\[9A - 12B + 16A + 12B = 6 + 12\]
\[ \Rightarrow 25A = 18\]
\[ \Rightarrow A = \frac{18}{25}\]
\[\text{ Putting value of A} = \frac{18}{25} \text{ in eq} \left( 2 \right)\text{ we get, }\]
\[3 \times \frac{18}{25} - 4B = 2\]
\[ \Rightarrow \frac{54}{25} - 2 = 4B\]
\[ \Rightarrow \frac{4}{25 \times 4} = B\]
\[ \Rightarrow B = \frac{1}{25}\]

Thus, substituting the values of A,B and C in eq (1) we get ,

\[I = \int\left[ \frac{\frac{18}{25}\left( 3 \sin x + 4 \cos x \right) + \frac{1}{25} \left( 3 \cos x - 4 \sin x \right)}{\left( 3 \sin x + 4 \cos x \right)} \right]dx\]
\[ = \frac{18}{25}\int dx + \frac{1}{25}\int\left( \frac{3 \cos x - 4 \sin x}{3 \sin x + 4 \cos x} \right)dx\]
\[\text{ Putting 3 sin x + 4 cos x = t}\]
\[ \Rightarrow \left( 3 \cos x - 4 \sin x \right) dx = dt\]
\[ \therefore I = \frac{18}{25}\int dx + \frac{1}{25}\int\frac{1}{t}dt\]
\[ = \frac{18x}{25} + \frac{1}{25} \text{ ln }\left| t \right| + C\]
\[ = \frac{18x}{25} + \frac{1}{25} \text{ ln }\left| 3 \sin x + 4 \cos x \right| + C\]